# If $f$ is an entire function for which $f^{-1}(B_r)$ is bounded for all $r >0$ then $\infty$ are removable singularity and pole

Let $$\mathbb{B}_r$$ denote the closed disk $$\{z \in \mathbb{C} : |z| \le r \}$$. State whether $$\infty$$ is a removable singularity , pole , or essential singularity in the following statement

$$f$$ is an entire function for which $$f^{-1}(B_r)$$ is bounded for all $$r >0$$

My attempt :Here $$\infty$$ are both removable singularity and pole

Inverse image of closed unit disk of $$f$$ is bounded so $$f$$ can not have essential singularity at $$\infty$$

Removable singularity :-

Suppose that $$f$$ is an entire function that has a removable singularity at infinity.Then there exist an entire function $$g$$ such that $$g(z)= f(1/z)$$ for all $$z \in \mathbb{B_r}-\{0\}$$.This implies that $$\lim_{z\to \infty} f(1/z)=f(0)$$ which in turn implies that $$g$$ is bounded.Since g is a bounded entire function, by liouville's theorem , $$g$$ is constant.Hence $$f$$ is constant $$\implies f^{-1}(B_r)$$ is bounded for all $$r >0$$

For pole

take $$f(z)= z$$ , $$g(z)=f(1/z)=1/z$$ where $$|z|\le r >0$$. $$g(z)$$ has pole at $$z=0 \implies f$$ has a pole at $$\infty$$ $$\implies f^{-1}(B_r)$$ is bounded for all $$r >0$$

Your second example is fine. For removable singularity the argument is simpler. Any entire function with removable singularty at $$\infty$$ is a constant by Liouville's Theorem because it is bounded.

If $$f$$ has an essential singularity at $$\infty$$ then it would take values in $$B_1$$ in every neighborhood of $$\infty$$ so $$f^{-1} (B_1)$$ is not bounded. Hence, $$f$$ cannot have an essential singularity at $$\infty$$.

• @wasiu If it is a removable singularity then $f$ is a consatnt. So $f^{-1}(B_r)=\mathbb C$ for large enough $r$ , contradicting the hypothesis. May 26, 2022 at 4:45
• okay got it ,thanks you @Kavi sir May 26, 2022 at 4:48